Thermodynamics filled in class notes_Part_28

Thermodynamics filled in class notes_Part_28 - 63 2.8....

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Unformatted text preview: 63 2.8. DIESEL kJ kg K T2 (K ) so 2 T kJ kg K −4 F (T2 ) 8.01581 8.57198 × 10 8.13493 8.97387 × 10−2 7.88514 −9.60091 × 10−2 7.95207 −4.64783 × 10−2 8.01237 −2.37449 × 10−3 900 1000 800 850 899.347 Table 2.3: Iteration for T2 . This then becomes a trial and error search in Table A7.1 for the T2 which satisfies the previous equation. If one guesses T2 ∼ 900 K , one gets so 2 = 8.01581 kJ/kg/K , and the right side evaluates to T 0.000857198 kJ/kg/K . Performing a trial and error procedure, one finds the results summarized in Table 2.3. Take T2 = 900 K as close enough! Then P2 T2 V1 , T1 V2 = P1 = (0.1 M P a) (2.420) 900 K (20), 288 K 6.25 M P a. = (2.421) (2.422) Now at state 2, T2 = 900 K , one finds that h2 = 933.15 kJ , kg u2 = 674.82 kJ . kg (2.423) The heat is added at constant pressure, so qH h3 = h3 − h2 , = h2 + qH , (2.424) (2.425) kJ kJ + 1800 kg kg kJ = 2733.15 . kg = 933.15 , (2.426) (2.427) One can then interpolate Table A7.1 to find T3 and so 3 T 2733.15 T3 = (2350 K ) + 2755 = so 3 T = kJ kg kJ kg − 2692.31 − 2692.31 kJ kg kJ kg ((2400 K ) − (2350 K )), (2.428) 2382.17 K, 9.16913 kJ kg K 2733.15 + 2755 = 9.18633 (2.429) kJ kg kJ kg kJ . kg K − 2692.31 − 2692.31 kJ kg kJ kg 9.19586 kJ kg K − 9.16913 kJ kg K , (2.430) (2.431) CC BY-NC-ND. 18 November 2011, J. M. Powers. 64 CHAPTER 2. CYCLE ANALYSIS T4 (K ) so 4 T 400 800 1400 1200 1300 1250 1288 kJ kg K kJ kg K −1 F (T4 ) −9.34561 × 10 −4.07614 × 10−1 7.55464 × 10−2 −6.31624 × 10−2 8.36535 × 10−3 −2.68183 × 10−2 −1.23117 × 10−4 7.15926 7.88514 8.52891 8.34596 8.44046 8.39402 8.42931 Table 2.4: Iteration for T4 . One also has P3 = P2 = 6.25 M P a. Now get the actual entropy at state 3: s3 (T3 , P3 ) = = = P3 , Pref kJ kJ − 0.287 9.18633 kg K kg K kJ 7.99954 . kg K so 3 − R ln T (2.432) ln 6.25 M P a , 0.1 M P a (2.433) (2.434) Now 3 → 4 is an isentropic expansion to state 4, which has the same volume as state 1; V1 = V4 . So the ideal gas law gives P4 V4 T4 P4 P1 P4 Pref P1 V1 , T1 T4 V1 , T1 V4 T4 V1 . T1 V4 = = = (2.435) (2.436) (2.437) Now consider the entropy at state 4, which must be the same as that at state 3: s4 (T4 , P4 ) = so 4 − R ln T s3 P4 , Pref (2.438) T4 V1 = so 4 − R ln T T1 V4 , (2.439) =1 kJ 7.99954 kg K = so 4 T 0 = so 4 T kJ − 0.287 kg K kJ − 0.287 kg K ln ln T4 288 K T4 288 K , − 7.99954 (2.440) kJ kg K = F (T4 ). (2.441) Using Table A7.1, this equation can be iterated until T4 is found. So T4 = 1288 K. At this temperature, one has kJ kJ , u4 = 1011.98 . (2.442) h4 = 1381.68 kg kg CC BY-NC-ND. 18 November 2011, J. M. Powers. ...
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